Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is:
1.`CH -= CH`
2.`CH_2 = CH_2`
3.`CH_3 - CH_3`
4.`CH_4`

Options

A. `CH -= CH`
B. `CH_2 = CH_2`
C. `CH_3 - CH_3`
D. `CH_4`

Answer

To determine which hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide that can be converted to a gaseous hydrocarbon containing less than four carbon atoms, we will analyze each option step by step.

### Step 1: Analyze Option 1 - `CH ≡ CH` (Ethyne)
- Ethyne reacts with bromine (Br2) to form a dibromide:
\[ \text{CH} \equiv \text{CH} + \text{Br}_2 \rightarrow \text{HC} = \text{CHBr} \]
- This results in an alkyl bromide, but it is not suitable for Wurtz reaction as it does not yield a hydrocarbon with less than four carbon atoms. Therefore, **Option 1 is eliminated**.

### Step 2: Analyze Option 2 - `CH2 = CH2` (Ethene)
- Ethene reacts with bromine to form a dibromide:
\[ \text{CH}_2 = \text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{BrCH}_2\text{Br} \]
- This reaction also forms a dibromide, which does not lead to a gaseous hydrocarbon with less than four carbon atoms. Thus, **Option 2 is eliminated**.

### Step 3: Analyze Option 3 - `CH3 - CH3` (Ethane)
- Ethane reacts with bromine to yield:
\[ \text{CH}_3 - \text{CH}_3 + \text{Br}_2 \rightarrow 2 \text{CH}_3\text{CH}_2\text{Br} \]
- The product is an alkyl bromide (brominated ethane). When subjected to Wurtz reaction:
\[ 2 \text{CH}_3\text{CH}_2\text{Br} + 2 \text{Na} \rightarrow \text{C}_4\text{H}_{10} + 2 \text{NaBr} \]
- This results in butane (C4H10), which contains four carbon atoms, not less than four. Therefore, **Option 3 is eliminated**.

### Step 4: Analyze Option 4 - `CH4` (Methane)
- Methane reacts with bromine to produce:
\[ \text{CH}_4 + \text{Br}_2 \rightarrow 2 \text{CH}_3\text{Br} \]
- The product is methyl bromide (CH3Br). In the Wurtz reaction:
\[ 2 \text{CH}_3\text{Br} + 2 \text{Na} \rightarrow \text{C}_2\text{H}_6 + 2 \text{NaBr} \]
- This results in ethane (C2H6), which contains less than four carbon atoms. Therefore, **Option 4 is the correct answer**.

### Final Answer:
The hydrocarbon (A) is **`CH4` (Methane)**.

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